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Why Do You Believe Superposition/entanglement Include Spacetime?


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There is no "boundary" between quantum and classical. It is just a matter of scale, specifically the size of quantities compared to Planck's Constant.   We understand very well how uncertainty princip

*pets polymath*  

Quantum mechanics has proven without any doubt, that particles are both particle-like and wave-like - it's only macroscopic systems in which the wavelike nature becomes extremely small, so much so it's almost impossible to detect. In other words, the wavelength of a particle diminishes in a busy environment.

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Spacetime as a dimension/fabric

I'm not making anything up, just pointing out the obvious. In what world are probabilities good enough? Look at the uncertainty principle, you can't know both momentum and position at the same time because the object is missing a dimension.

Probabilities are good enough in the QM world, evidently, and that is the real world if you are a chemist or a particle physicist.

 

 

In QM the position and momentum operators do not commute. Or to put it another way, position and momentum distributions are Fourier transforms of one another. Nothing to do with anything "missing a dimension", so far as I can see. 

Edited by exchemist
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They aren't both at the same time. You really think matter goes through both slits of the double slit experiment and interferes with itself?

Yes. Why not? People seem to forget that the result of the double slit experiment is exactly what QM predicted it would be. So why do they act surprised?  

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They aren't both at the same time. You really think matter goes through both slits of the double slit experiment and interferes with itself?

 

Here is some interpretations:

 

1. The particle can be both a wave and a particle

 

or

 

2. They are either particle and behaving like a wave sometimes

 

Do not mistake that the first statement may be in contradiction with the second.... so why? On one hand, we have experimental evidence that the wave hits a detector and appears on the detector as a point - suggesting a particle nature. However, because of a single wave diffraction, it also acts like a wave.... do they both need to be in contradiction? The answer is no, and deBroglie hit on this, stating that a particle may be guided by its own wave. 

 

DeBroglie realized this would predict what is called an empty wave... so I took his idea further. First, what do we mean by an empty wave? Well.... consider for a moment that the particle is only a wave, then it has to travel through both slits - however, if it is a wave which guides the particle through the slits, then a particle can only travel through one slit, this predicts that there is such a thing as an empty wave - a wave moving through space without any energy or mass.

 

I came to realize, this could only be possible if the particle was ''in phase'' with the empty wave which has to include some type of quantum entanglement (or what we call a correlation). So, with that idea in mind, I decided to take the standard quantum entangle equations and applied it to the notion of an occupied space and a non-occupied state.

 

EMPTY WAVE CORRELATION WITH OCCUPIED WAVE

 

There's loads of different approaches I could take to describe how the particle becomes entangled with the empty wave, but sometimes a logical argument alone can suffice. Below, I use some notation some people may not be so acquainted with so I give it some explanation. It does involve the concepts of entanglement.

 

[math]H(\Psi) = H(\psi_{empty}|\psi_{occup}) + H(\psi_{empty} : \psi_{occup})[/math]

 

Here, [math]H(\psi_{empty}|\psi_{occup})[/math] is the entropy in [math]\psi_{empty}[/math] (after) having measured the system that became correlated in the occupied wave function [math]\psi_{occup}[/math].

 

Further, the notation [math]H (\psi_{empty} : \psi_{occup})[/math] is the information gained about [math]\psi_{empty}[/math] by measuring the occupied state [math]\psi_{occup}[/math]. This reveals a conservation in the second law not so dissimilar to the Shannon entropy in terms of information theory. Likewise from information theory, we can talk about the empty wave and occupied wave in terms of the upper bound of correlation which is given as:

 

[math]H(\Psi) = H(\rho_{\psi_{empty}}) + H(\rho_{\psi_{occup}}) - H(\rho_{\psi_{empty,\ occup}})[/math]

 

Which is a [quantum discord] attempt to unify an empty wave with an occupied wave. Let's talk a bit about correlation in terms of the quantum discord.

 

Quantum Correlations

 

When the systems are correlated we mean there exists a quantum entanglement - an entangled state is a linear superposition of separable states; the quantum divergence is

 

[math]S(A:B ) \leq 2\min[s(A),S(B )][/math]

 

Unlike the classical conditional entropy,

 

[math]S(a|b ) = S(a,b ) - S(b )[/math]

 

which remains positive always, while the quantum mechanical equivalent form

 

[math]S(\rho_A|\rho_B ) = S(\rho_{AB }) - S(\rho_B )[/math]

 

is not. The state is entangled if [math]S(\rho_A|\rho_B ) < 0[/math] where

 

[math]S(\rho_A \otimes \rho_B ) = S(\rho_A) + S(\rho_B )[/math]

 

Note also that

 

[math]S(\rho_A \otimes \rho_B|\rho_{AB}) = S(\rho_A) + S(\rho_B ) - S(\rho_{AB}) \geq \frac{1}{2}| \rho_A \otimes \rho_B - \rho_{AB}|^2[/math]

 

[math]\geq \frac{(Tr(\mathcal{O}_A \mathcal{O}_B )( \rho_A \otimes \rho_B - \rho_{AB} ))^2}{2|\mathcal{O}_A|^2|\mathcal{O}_B|^2}[/math]

 

[math]= \frac{( <\mathcal{O}_A><\mathcal{O}_B> - <\mathcal{O}_A\mathcal{O}_B>)^2}{2|\mathcal{O}_A|^2|\mathcal{O}_B|^2}[/math]

 

Which is known as the upper bound of correlation. In which the mutual information is

 

[math]I(A:B ) = I(\rho_A) + I(\rho_B ) - I(\rho_{AB})[/math]

 

Quantum discord - Wikipedia

Edited by Dubbelosix
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I know the math works for all the interpretations. Aren't you more interested in what is really happening?

 

I find this a really strange question, of course I am interested in what is going on, why do you think I have been taking the time to talk about interpretations? Why do you think I entertain various theories?

 

Please don't get back-chatty with ''I know the math works,'' because I seriously doubt it - even if you did, your following statements are totally bizarre to me.

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There isn't a reason to entertain other interpretations if you were to believe what I've said. I stumbled on this obvious thing about QM/Spacetime and it's been squeezing water from a rock to get anyone see what I'm pointing at. Everyone has been programmed to not ask what the unobservable is and to just do the math. 

 

You can have QM without spacetime just fine, but spacetime needs QM as a base.

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There isn't a reason to entertain other interpretations if you were to believe what I've said.

 

That's the problem though, I don't believe what you have said because I know better... tell me... when will your next account be created to try and deceive me?? I know who you are, I am just curious as to why you think I am so stupid I don't know these things about your monikers. I think from now on, I won't acquiesce or entertain this any more.

 

You are really testing my patience, do you want me to name all your monikers here at the site? Continuing to create new accounts, trying to word things differently, is one the most disingenuous things I have ever seen. I too made accounts on other sites, but that was to by-pass the one account, you have a whole list of accounts... also, stop bothering me in my private messages.

Edited by Dubbelosix
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